Suppose that the number of tourists (random variable X) using Metro to get to th

Question

Suppose that the number of tourists (random variable X) using Metro to get to the National Mall on any given day during the summer is normally distributed with mu = 49000 and sigma = 9000. (remember to show your work!) (Table V on last page) What is the probability of less than 46, 500 tourists using Metro to get to the Mall on a randomly selected summer day? b) Metro has discovered that there are train capacity issues when the number of tourists going to the Mall exceeds 62,000. What is the probability of that occurring a randomly selected summer day? c) Metro decides that they are going to adjust their summer schedules to have adequate capacity to handle the summer tourists 92% of the time. What number of tourists would be at the 92^nd percentile?

Explanation / Answer

Mean ( u ) =49000
Standard Deviation ( sd )=9000
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P(X < 46500) = (46500-49000)/9000
= -2500/9000= -0.2778
= P ( Z <-0.2778) From Standard Normal Table
= 0.3906                  
b.
P(X > 62000) = (62000-49000)/9000
= 13000/9000 = 1.4444
= P ( Z >1.444) From Standard Normal Table
= 0.0743                  
c.
P ( Z < x ) = 0.92
Value of z to the cumulative probability of 0.92 from normal table is 1.405
P( x-u/s.d < x - 49000/9000 ) = 0.92
That is, ( x - 49000/9000 ) = 1.41
--> x = 1.41 * 9000 + 49000 = 61645.644                  

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.